Question

Prove that integral of `sqrt(x^2+a^2)` is `x/2×(x^2+a^2)^(-1/2)+a^2/2×log|x+(x^2+a^2)^(-1/2)|+c`?

Answer 1

`int sqrt(x^2+a^2)*dx`

=`x/2*sqrt(x^2+a^2)+a^2/2*Ln(x+sqrt(x^2+a^2))+C`

Explanation:

`int sqrt(x^2+a^2)*dx`

After using `x=asinhu` and `dx=acoshu*du` transforms, this integral became

`int a^2*(coshu)^2*du`

=`a^2/2int (1+cos2hu)*du`

=`a^2/2*(u+1/2*sin2hu)+C_1`

=`a^2/2*u+a^2/4*sin2hu+C_1`

=`a^2/2*u+a^2/4*2sinhu*coshu+C_1`

=`a^2/2*u+a^2/2*sinhu*coshu+C_1`

After using `x=sinhu`, `sinhu=x/a`, u=`arcsin(x/a)=Ln((x+sqrt(x^2+a^2))/a) and`coshu=sqrt(x^2+a^2)/a# inverse transforms, I found

`int sqrt(x^2+a^2)*dx`

=`x/2*sqrt(x^2+a^2)+a^2/2*Ln((x+sqrt(x^2+a^2))/a)+C_1`

=`x/2*sqrt(x^2+a^2)+a^2/2*Ln(x+sqrt(x^2+a^2))+C`

Note: `C=C_1-Lna`

Answer 2

Please see a Proof in the Explanation.

Explanation:

Prerequisites : Integration by Parts (IBP) :

`intu*vdx=uintvdx-int{(du)/dx*intvdx}dx`.

Let, `I=intsqrt(x^2+a^2)dx=intsqrt(x^2+a^2)*1dx`.

We take :

`u=sqrt(x^2+a^2):.(du)/dx=1/(2sqrt(x^2+a^2))*2x=x/sqrt(x^2+a^2)`.

`v=1 rArr intvdx=x`.

`:.I=xsqrt(x^2+a^2)-int{x/sqrt(x^2+a^2)*x}dx`,

`=xsqrt(x^2+a^2)-int{x^2/sqrt(x^2+a^2)}dx`,

`=xsqrt(x^2+a^2)-int{(x^2+a^2)-a^2}/sqrt(x^2+a^2)}dx`,

`=xsqrt(x^2+a^2)-int{(x^2+a^2)/sqrt(x^2+a^2)-a^2/sqrt(x^2+a^2)}dx`,

`=xsqrt(x^2+a^2)-intsqrt(x^2+a^2)dx+a^2int1/sqrt(x^2+a^2)dx, or,`,

`I=xsqrt(x^2+a^2)-I+a^2ln|x+sqrt(x^2+a^2)|, i.e.,`

`2I=xsqrt(x^2+a^2)+a^2ln|x+sqrt(x^2+a^2)|`.

`rArr I=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+C`.